# Taylor's theorem

In calculus, Taylor's theorem, named after the mathematician Brook Taylor, who stated it in 1712, gives the approximation of a differentiable function near a point by a polynomial whose coefficients depend only on the derivatives of the function at that point. This result was first discovered 41 years earlier in 1671 by James Gregory.

## Taylor's theorem in one variable

The most basic example of Taylor's theorem is the approximation of the exponential function $\displaystyle \textrm{e}^x$ near x = 0. Namely,

$\displaystyle \textrm{e}^x \approx 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots + \frac{x^n}{n!}.$

The precise statement of the theorem is as follows: If n ≥ 0 is an integer and f is a function which is n times continuously differentiable on the closed interval [a, x] and n + 1 times differentiable on the open interval (a, x), then we have

$\displaystyle f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \frac{f^{(2)}(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n + R_n$

Here, n! denotes the factorial of n, and Rn is a remainder term which depends on x and is small if x is close enough to a. Several expressions for Rn are available.

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The exponential function y = ex (red) and its corresponding Taylor's polynomial of degree 4 (blue)

The Lagrange form of the remainder term states that there exists a number ξ between a and x such that

$\displaystyle R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}.$

This exposes Taylor's theorem as a generalization of the mean value theorem. In fact, the mean value theorem is used to prove Taylor's theorem with the Lagrange remainder term.

The Cauchy form of the remainder term is

$\displaystyle R_n(x) = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt.$

This shows the theorem to be a generalization of the fundamental theorem of calculus.

For some functions f(x), one can show that the remainder term Rn approaches zero as n approaches ∞; those functions can be expressed as a Taylor series in a neighbourhood of the point a and are called analytic.

Taylor's theorem (with the integral formulation of the remainder term) is also valid if the function f has complex values or vector values. Furthermore, there is a version of Taylor's theorem for functions in several variables.

For complex functions analytic in a region containing a circle C surrounding a and its interior, we have a contour integral expression for the remainder

$\displaystyle R_n(x) = \frac{1}{2 \pi i}\int_C \frac{f(z)}{(z-a)^{n+1}(z-x)}dz$

valid inside of C.

## Taylor's theorem for several variables

Using multi-index notation (see also Taylor series in several variables), Taylor's theorem can be generalized to several variables as follows. Let B be a ball in RN centered at a point a, and f be a real-valued function defined on the closure $\displaystyle \bar{B}$ having n+1 continuous partial derivatives at every point. Taylor's theorem asserts that for any $\displaystyle x\in B$ ,

$\displaystyle f(x)=\sum_{|\alpha|=0}^n\frac{D^\alpha f(a)}{\alpha!}(x-a)^\alpha+\sum_{|\alpha|=n+1}R_{\alpha}(x)(x-a)^\alpha$

where the summation extends over multi-indices α.

The remainder terms satisfy the inequality

$\displaystyle |R_{\alpha}(x)|\le\sup_{y\in\bar{B} }\left|\frac{D^\alpha f(y)}{\alpha!}\right|$

for all α with |α|=n+1. As was the case with one variable, the remainder terms can be described explicitly. See the proof for details.

## Proof: Taylor's theorem in one variable

We first prove Taylor's theorem with the integral remainder term.

The fundamental theorem of calculus states that

$\displaystyle f(x) = f(a) + \int_a^x (x-t)^0 \, f'(t) \, dt.$

This proves the theorem for n = 0.

Integration by parts yields the case n = 1:

$\displaystyle f(x) = f(a) +f'(a)\,(x-a)+\int_a^x (x-t)^1 \, f''(t) \, dt.$

By repeating this process, we may derive Taylor's theorem for higher values of n.

This can be formalized by applying the technique of induction. So, suppose that Taylor's theorem holds for a particular n, that is, suppose that

$\displaystyle f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n + \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt. \qquad(*)$

We can again rewrite the integral using integration by parts. An antiderivative of (x − t)n as a function of t is given by −(xt)n+1 / (n + 1), so

$\displaystyle \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt$
$\displaystyle {} = - \left[ \frac{f^{(n+1)} (t)}{(n+1)n!} (x - t)^{n+1} \right]_a^x + \int_a^x \frac{f^{(n+2)} (t)}{(n+1)n!} (x - t)^{n+1} \, dt$
$\displaystyle {} = \frac{f^{(n+1)} (a)}{(n+1)!} (x - a)^{n+1} + \int_a^x \frac{f^{(n+2)} (t)}{(n+1)!} (x - t)^{n+1} \, dt.$

Substituting this in (*) proves Taylor's theorem for n + 1, and hence for all nonnegative integers n.

The remainder term in the Lagrange form can be derived by the mean value theorem in the following way:

$\displaystyle R_n = \int_a^x \frac{f^{(n+1)} (t)}{n!} (x - t)^n \, dt =f^{(n+1)}(\xi) \int_a^x \frac{(x - t)^n }{n!} \, dt.$

The last integral can be solved immediately, which leads to

$\displaystyle R_n = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x-a)^{n+1}.$

## Proof: several variables

Let x=(x1,...,xN) lie in the ball B with center a. Parametrize the line segment between a and x by u(t)=a+t(x-a). We apply the one-variable version of Taylor's theorem to the function f(u(t)):

$\displaystyle f(x)=f(u(1))=f(a)+\sum_{i=0}^n\left.\frac{1}{i!}\frac{d^i}{dt^i}\right|_{t=0}f(u(t))\ +\ \int_0^1 \left. \frac{1}{(n+1)!}\frac{d^{n+1}}{ds^{n+1}}\right|_{s=0} f(u(s))ds.$

By the chain rule for several variables,

$\displaystyle \frac{d^i}{dt^i}f(a+t(x-a))=\sum_{|\alpha|=i}\left(\begin{matrix}i\\ \alpha\end{matrix}\right)(D^\alpha f)(a+t(x-a))$

where $\displaystyle \left(\begin{matrix}i\\ \alpha\end{matrix}\right)$ is the multinomial coefficient for the multi-index α. Since $\displaystyle \frac{1}{i!}\left(\begin{matrix}i\\ \alpha\end{matrix}\right)=\frac{1}{\alpha!}$ , we get

$\displaystyle f(x)= f(a)+\sum_{|\alpha|=0}^n\frac{1}{\alpha!}D^\alpha f(a)+\sum_{|\alpha|=n+1}\frac{(x-a)^\alpha}{\alpha!}\int_0^1 D^\alpha f(a+s(x-a))ds.$

The remainder term is given by

$\displaystyle \sum_{|\alpha|=n+1}\frac{(x-a)^\alpha}{\alpha!}\int_0^1 D^\alpha f(a+s(x-a))ds,$

The terms of this summation are explicit forms for the Rα in the statement of the theorem. These are easily seen to satisfy the required estimate.